# Is a Radon measure always positive on non-empty open sets? The same question about the Haar measure.

A measure is locally finite if it is finite on all compact sets from the underlying $$\sigma$$-algebra.

A measure is regular if every measurable set $$A\in\Sigma$$ can be approximated from above by open measurable sets: $$\mu(A)\,=\,\operatorname{inf}\,\left\{\,\mu(O)\;\Big{|}\;A\subseteq O\;,\;\;O\in\Sigma\,,\;\,O\;\,\mbox{open} \,\right\}\;\;,\;\;$$ and from below by compact measurable sets: $$\mu(A)\,=\,\operatorname{sup}\,\left\{\,\mu(C)\;\Big{|}\;C\subseteq A\,,\;\,C\in\Sigma\,,\;\,C\;\,\mbox{compact}\,\right\}\;\;.$$

A Radon measure is a Borel measure that is introduced on the Borel algebra of a Hausdorff topological space, and is regular and locally finite.

Now, my question: $$\;\;$$is a non-zero Radon measure always positive on non-empty open sets?

• you should specify that $\mu$ be non-zero (otherwise the answer is trivially no) Jan 17, 2021 at 2:48

Unless I'm overlooking something trivial, the answer is no. Look at the "restriction" of Lebesgue measure $$\lambda$$ on $$\Bbb{R}$$ to $$[0,1]$$, by which I mean $$\mu(E):= \lambda(E\cap [0,1])$$. Then, $$\mu$$ is a finite Borel measure on $$\Bbb{R}$$, hence is regular, but the open set $$(10,11)$$ has measure $$0$$ with respect to $$\mu$$.

• Thank you. And the Haar measure? Jan 17, 2021 at 3:01
• @Michael_1812 I actually don't know much about Haar measures, so I'm not the right person to ask lol (if you want, update your question, and maybe someone else can answer) Jan 17, 2021 at 3:03
• Thank you. Will do. But first will cogitate on this a bit. Jan 17, 2021 at 3:13

As mentioned in the comments, you should require $$\mu \ne 0$$. But even then, there are trivial counterexamples.

For example, let $$U$$ be a non-empty subset of any locally compact Hausdorff space $$X$$ and $$x \notin U$$. Then consider the Dirac measure $$\delta_x$$, which is Radon, and note that $$\delta_x(U) = 0.$$

• Yes, that's an eloquent counter-example. Thank you. Feb 5 at 16:33
• Could you please answer my question also for the Haar measure? Feb 5 at 16:34
• @Michael_1812 For the Haar measure the assertion is true, there is no counterexample. Feb 5 at 20:40
• Thank you! Do you happen to know a simple proof or a reference? Feb 6 at 4:26
• @Michael_1812 It is in Folland's book "Real analysis" in the chapter about Haar measure. Feb 6 at 7:38

Suppose there is an open set $$\sigma_0$$ such that $$\mu(\sigma_0)=0$$. For an arbitrary open set $$\sigma$$, consider the infinite union $$U=\bigcup_{x\in\sigma,\,y\in\sigma_0} g(y,\,x)\,\sigma_0~\,,$$ where $$g(y,\,x)$$ is the group element sending a point $$y\in\sigma_0$$ to a point $$x\in\sigma$$. Since $$x$$ ranges over the entire $$\sigma$$, we observe that $$\sigma\in U~.$$ By definition of topological group, $$g(y,\,x)\sigma_0$$ is open for $$\forall x,\,y$$. For a Haar measure, $$\mu(\, g(y,\,x)\sigma_0\,)=\mu(\sigma_0)=0$$, whence $$\mu(U)=0$$. By definition of topological space, $$U$$ is open. We then observe that $$\sigma\in U$$ and $$\mu(U)=0$$ together yield $$\mu(\sigma)=0$$. Thus the existence of one measure-zero open set $$\sigma_0$$ ensures that all open sets are of zero measure. Among these, is our entire topological space. So $$\mu$$ is a zero norm, which contradicts the initial assumption.